On the Hilbert Functions of Disjoint Unions of a Linear Space and Many Lines in P n

International Mathematical Forum, 5, 2010, no. 16, 787-798 On the Hilbert Functions of Disjoint Unions of a Linear Space and Many Lines in P n E. Ballico 1 Dept. of Mathematics University of Trento 38123 Povo (TN), Italy ballico@science.unitn.it Abstract. Let A P n, n m +2 4, be an m-dimensional linear subspace. We show the existence of schemes X = A B P n with B disjoint union of lines and a certain prescribed Hilbert functions (not the general ones, but near the best ones). Mathematics Subject Classification: 14N05 Keywords: postulation: Hilbert function; unions of planes; unions of lines 1. Introduction In [3] and [4] E. Carlini, M. V. Catalisano and A. V. Geramita made a conjecture about the Hilbert function of general disjoint unions of linear spaces: their Hilbert function should be the minimal one compatible with the Hilbert function of each linear space and of the ambient projective space. Here we consider a different, but related problem: the construction of disjoint unions of linear subspaces of P n whose Hilbert function is not the best possible for their numerical data (bipolynomial Hilbert function in the sense of [3], [4]), but that its failure is completely understood. We first prove the following result (the bipolynomial case) in which no failure arises. Proposition 1. Fix integers n, m, y such that n m +2 4 and y 0. Let X P n be a general union of an m-dimensional linear subspace and of y lines. Let k be the minimal positive integer such that m + k n + k (1) +(k +1)y m n Then h 0 (I X (k 1)) = 0 and h 1 (I X (t)) = 0 for all t k. 1 The author was partially supported by MIUR and GNSAGA of INdAM (Italy).

788 E. Ballico The case m = 2 of Proposition 1 is contained in [4], Theorem 5.1. The integer k appearing in the statement of Proposition 1 (and below in Theorems 1 and 2) is called the critical value of X or of the triple (n, m, y). Castelnuovo-Mumford s lemma gives that if h 1 (I X (k)) = 0, then h 1 (I X (t)) = 0 for all t > k. In the terminology of [3] and [4] the scheme X as in the statement of Proposition 1 has the bipolynomial Hilbert function h X, i.e. h X (z) = min{ ( n+z n, m+z ) m +(m +1)y} for all z N. We will often say that X has maximal rank or the expected postulation. Now we come to our main contribution here. Theorem 1. Fix integers n, m, y such that n m +2 4 and y>n+2. Let k be the minimal positive integer such that (1) is satisfied. Fix an integer s such that k s y n. Then there exists a disjoint union X P n of an m-dimensional linear space and y disjoint lines such that h 1 (I X (t)) = 0 for all t s +1 and h 1 (I X (t)) = s +1 t for all k t s. Theorem 2. Fix an integer e > 0. There is an integer α e 0 with the following properties. Fix positive integers n, m, y such that n m +2 4. Let k be the minimal positive integer such that (1) is satified. Assume k α e. and m + k 1 n + k 1 (2) + ky + k 2 (3e +2k) m n Then there exists a disjoint union X P n of an m-dimensional linear space and y disjoint lines such that : (i) h 1 (I X (t)) = 0 for all t k +1, h 1 (I X (k)) = e and h 0 (I X (k 1)) = 0; (ii) there are e disjoint lines L i P n, 1 i e, such that (X L i )=k+2 for all i and h 1 (I X (L1 L e)(k)) = e. We will find a solution X of the statement of Theorem 1 and a line L P n such that (X L) =s +2. Let Y P n be any disjoint union of linear subspaces. Assume the existence of a line L P n such that L Y and (L Y ) s + 2. The line L shows that the homogeneous ideal of Y is not generated by forms of degree at most s + 1. Thus Castelnuovo-Mumford s lemma gives h 1 (I Y (s)) > 0. Since h i (O Y (t)) = 0 for all t>0 and all i>0, i.e. h i+1 (I Y (t)) = 0 for all t>0 and all i > 0, a standard exact sequence gives that the function N N defined by t h 1 (I Y (t)) is strictly decreasing, until it goes to zero. Thus h 1 (I Y (t)) s +1 t for every non-negative integer t s. This observation explains the power of the statement of Theorem 1. For Theorem 2 we will use e (k + 2)-secant lines added in the last step of our inductive proof. The case (n, m) =(3, 1) was omitted from the statement of Theorem 2, because a more explicit form of it is known ([1], Proposition 1.1). We work over an algebraically closed field K such that char(k) = 0. We use the characteristic zero assumption to quote two key observations (Remarks 4 and 5) which we lift from [2].

Unions of linear spaces 789 2. Preliminaries Remark 1. Fix an integer s>0. Let X P n be a projective scheme such that h i (X, O X (s i 1)) = 0 for all i 1, i.e. for all integers i such that 1 i dim(x). Assume the existence of a line D P n such that length(x D) =s. In particular we are assuming D X. Since length(x D) =s, the homogeneous ideal of X cannot be generated in degree s 1. Hence Castelnuovo-Mumford s lemma gives h 1 (I X (s 2)) 0. For any integral scheme M P m, m 2, and any P M reg let χ M (P ) denote the first infinitesimal neighborhood of P in M, i.e. the closed subscheme of M with (I P,M ) 2 as its ideal sheaf. Thus χ M (P ) red = {P }, length(χ M (P )) = dim P (M) + 1 and χ M (P )=χ TP M(P ). If dim P (M) = 1 (resp. dim P (M) =2) we often call χ M (P )atangent vector (resp. a planar length 3 subscheme) of P m. The set of all tangent vectors (resp. length 3 planar subschemes) of P m is parametrized by an integral quasi-projective variety. Hence for all non-negative integers α, β we may look at the cohomological properties of a general union Z P m of α tangent vectors and β length 3 planar subschemes of P m. Remark 2. Fix a reducible conic D P n, n 3, and let P be its singular point. Let M P n be any 3-dimensional linear space containing D. The scheme D χ M (P ) is a flat degeneration inside M and hence inside P n of a flat family whose general element is the disjoint union of 2 lines ([5]). Let H P n be any hyperplane containing P, but no irreducible component of D. The scheme D H is a tangent vector of H with P as its reduction. Fix P and H, but take as D a general reducible conic with P as its singular point. Then D H is the general tangent vector of H with P as its support. Now take M general. Then (D χ M (P )) H = χ M H (P ). Thus (D χ M (P )) H is a general length 3 planar subscheme of H with P as its reduction. Remark 3. Let X be any projective scheme and D any effective Cartier divisor of X. For any closed subscheme Z of X let Res D (Z) denote the residual scheme of Z with respect to D, i.e. the closed subscheme of X with I Z : I D as its ideal sheaf. For every L Pic(X) we have the exact sequence (3) From (3) we get 0 I ResD (Z) L( D) I Z L I Z D,D (L D) 0 h i (X, I Z L) h i (X, I ResD (Z) L( D)) + hi (D, I Z D,D (L D)) for every integer i 0. If h 2 (X, I ResD (Z) L( D)), then h1 (X, I Z L) h 1 (D, I Z D,D (L D)). If h 2 (X, I ResD (Z) L( D)) = h0 (D, I Z D,D (L D)) = 0, then h 1 (X, I Z L) =h 1 (X, I ResD (Z) L( D)) + h1 (D, I Z D,D (L D)). We borrow the following two observations from [2].

790 E. Ballico Remark 4. Fix any scheme W P m and any integer t 0. Then (after fixing W and t) fix a general P P m and a general tangent vector v such that v red = {P }. Since P is general, h 0 (I W {P } (t)) = max{0,h 0 (I W (t)) 1}. Thus h 1 (I W {P } (t)) = h 1 (I W (t)) if h 0 (I W (t)) > 0 and h 1 (I W {P } (t)) = h 1 (I W (t))+1 if h 0 (I W (t)) = 0. Since v red = {P } and length(v) = 2, we have h 0 (I W {P } (t)) 1 h 0 (I W v (t)) h 0 (I W {P } (t)). Hence h 0 (I W v (t)) = 0 if h 0 (I W (t)) 1. Now assume h 0 (I W (t)) 2. Thus the rational map η from P m into P N, N := h 0 (I W (t)) 1, is not constant. Since we chosed P general after fixing W and t, P is not a base point of η and the differential d P η of η at P has positive rank (here we use the characteristic zero assumption). Since v is a general element of the tangent space T P P m, the kernel of the linear map d P η does not contain v. Thush 0 (I W v (t)) = h 0 (I W (t)) 2, i.e. h 1 (I W v (t)) = h 1 (I W (t)). Remark 5. Fix any scheme W P m, m 2, and any integer t>0. Assume h 0 (I W (t)) 3. Then (after fixing W and t) fix a general lenght 3 planar subscheme Z of P m. Since Z contains a general tangent vector of P m, Remark 4 gives h 0 (I W Z (t)) h 0 (I W (t)) 2. Since Z has length 3, h 0 (I W Z (t)) h 0 (I W (t)) 3. Assume h 0 (I W Z (t)) = h 0 (I W (t)) 2, i.e. assume h 1 (I W Z (t)) >h 1 (I W (t)). Let Ψ denote the rational map induced by the linear system I W (t). Our assumption is equivalent to having Ψ rank 1 at a general point of P m. Since we are in characteristic zero, the image of the rational map Ψ must be a curve. Hence for a general P 1 P m the closure of the fiber of Ψ passing through P 1 contains a hypersurface Σ 1. Since h 0 (I W Z {P1 }(t)) = h 0 (I W Z (t)) 1, if h 0 (I W (t)) 4 we may continue. Take a general P 2 P m \Σ 1. We get a hypersurface Σ 2 such that P 2 Σ 2 and h 0 (I W Z Σ1 Σ 2 (t)) = h 0 (I W Z {P1,P 2 }(t)) = h 0 (I W Z (t)) 2. And so on. Since no degree t hypersurface may contain t + 1 distinct hypersurfaces, this is absurd if h 0 (I W (t)) t + 2. If for some reason we may exclude the case deg(σ 1 ) = 1, it would even be sufficient to check (or assume) the inequality h 0 (I W (t)) (t +2)/2. 3. The proofs We fix a hyperplane H of P n (or P r ) and an m-dimensional (resp. t- dimensional) linear subspace A of P n (or P r ) not contained in H. For all integers r, t, k such that r t +2 4 and k 1 we define the integers a r,k and b r,k by the relations t + k r + k (4) +(k +1) a r,t,k b r,t,k =, 0 b r,t,k k t r Taking the difference of the equation in (4) and the same equation for the triple (r,t,k )=(r, t, k 1) we get: t + k 1 k + r 1 (5) +(k + 1)(a r,t,k a r,t,k 1 )+b r,t,k 1 b r,t,k = t 1 r 1

Unions of linear spaces 791 For all integers r, t, k such that r t +2 4 and k 1 we define the following Assertion A r,t,k r t +2. Assertion A r,t,k, r t +2 4: Let X P r be a general union of the t-dimensional linear subspace A, a r,t,k 2b r,t,k disjoint lines and b r,t,k reducible conics with their singular points contained in H. Then h i (I X (k))=0,i =0, 1. In the set-up of Assertion A r,t,k we usually write A B D or A B D instead of X, where A is the t-dimensional linear space, B is a disjoint union of lines, D is a disjoint union of reducible conics with their singular point contained in H and A B = A D = B D =. Lemma 1. The Assertion A r,t,k is true for all integers r, t, k such that r t +2 4 and k 1. Proof. Notice that a r,t,1 2b r,t,1 (Remark 6), i.e. A r,t,1 is well-defined. Obvious properties of general disjoint linear spaces and planes spanned by a reducible conic give A r,t,1. Hence from now on we assume k 2. At each step we will silently use induction on k. Until step (d) we assume k 3. (a) Here we assume (r, t) =(4, 2). Fix a general solution Y = A B D of A 4,2,k 1. (a1) Here we assume b 4,2,k 1 b 4,2,k. Lemma 6 gives a 4,2,k a 4,2,k 1 b 4,2,k b 4,2,k 1. Lemma 5 gives a 4,2,k 1 2b 4,2,k 1 b 4,2,k b 4,2,k 1 (here we use k 2). Let E H be a general union of a 4,2,k 1 a 4,2,k lines, with the only restriction that exactly b 4,2,k b 4,2,k 1 of them contain a different point of B H. Set X := Y E. Thus X is a disjoint union of the plane A, a 4,2,k 2b 4,2k lines and b 4,2,k with reducible tangent vector with their singular points contained in H. Since Res H (X) =Y, we have h i (I ResH (k 1)) = 0, (X) i =0, 1. Since Y is general, H X is a general union of 1 + a 4,2,k a 4,2,k 1 lines, b 4,2,k 1 tangent vectors and a 4,2,k 1 2b 4,2.k 1 (b 4,2,k b 4,2,k 1 ) points. From (5) with (r, t) =(4, 2) we get (6) k +3 (k + 1)(1 + a 4,2,k a 4,2,k 1 )+2b 4,2,k 1 +(a 4,2,k 1 b 4,2.k 1 b 4,2,k )= 3 Since a general union of lines in H has maximal rank ([5]), Lemma 4 and (6) give h i (H, I X H (k)) = 0, i =0, 1. Apply Remark 3. (a2) Here we assume b 4,2,k 1 <b 4,2,k. Fix S Sing(D) such that (S) = b 4,2,k b 4,2,k 1. For each P S fix a 3-dimensional linear subspace H P of P 4 containing the connected component of D with P as its singular point. Set χ := P S χ HP (P ). Let E H be a general union of a r,2,k a r,2,k 1 lines. Set X := Y χ E. The scheme X is a flat limit of a flat family of disjoint unions of one plane, a r,2,k 2b r,2,k lines and b r,2,k reducible conics with their singular points contained in H (Remark 2). Since Res H (X) =Y, it is sufficient to prove h i (H, I X H (k))=0,i =0, 1 (Remark 3). Since X H is a general union of 1+a 4,2,k a 4,2,k 1 lines, b 4,2,k tangent vectors and b 4,2,k 1 b 4,2,k length 3 planar subschemes, we have h 0 (X H, O X H (k)) = k+3 3 and h 1 (X H, O X H (k))=0

792 E. Ballico for all i>0. Since a general union of 1+a 4,2,k a 4,2,k 1 lines have maximal rank ([5]), to conclude it is sufficient to apply first b 4,2,k 1 b 4,2,k times Remark 5 and then Remark 4. A warning: each time we apply Remark 5 we need to check a numerical inequality. In our case the numerical inequality to apply it the z-th times is that the length of the union of the zero-dimensional subschemes of X H with length 1 and b 4,2,k 1 b 4,2,k z + 1 of the length 3 planar subschemes is at least k + 2, i.e. we need a r,2,k 1 2b r,2,k 1 +2(b r,2,k 1 b r,2,k + 3(b 4,2,k 1 b 4,2,k z+1) k+2. The worst case is in the last application, i.e. if z = b 4,2,k 1 b 4,2,k 1. In this case we need the inequality a r,2,k 1 b r,2,k k 1, which is true by Lemma 7. (b) Here we assume t = 2 and r 5. We use induction on r, the starting point of the induction being true by step (a). However, here and in step (c) we have two options for the induction: the 2-dimensional (resp. t-dimensional) linear space may be added in H at the last step, applying [5] except at the very last step or we we may take it at the first step with the critical value k = 1. In all cases we use the second strategy. By the inductive assumption on k we may assume that A r,2,k 1 is true. Let Y = A B D be a general solution of A r,2,k 1. As in step (a) we distinguish the case b r,2,k 1 b r,2,k and the case b r,2,k 1 >b r,2,k. In both cases we make the same construction as in step (a). If b r,2,k 1 b r,2,k, then we take a general union E H of a r,2,k 1 a r,2,k lines, with the only restriction that exactly b r,2,k b r,2,k 1 of them contain a different point of B H. Here we use Lemmas 5 and 6 to get a r,2,k 1 2b r,2,k 1 b r,2,k b r,2,k 1 and a r,2,k 1 a r,2,k b r,2,k b r,2,k 1 (these inequalities are necessary to define E). Then we apply Remark 4. Now assume b r,2,k 1 <b 4,2,k. Fix S Sing(D) such that (S) =b r,2,k b r,2,k 1. For each P S fix a 3-dimensional linear subspace H P of P r containing the connected component of D with P as its singular point. (c) Here we assume t 3. We first do the case r = t + 2 and then by induction on r the general case. By steps (a) and (b) we may assume that A r,t,k is true for all (r,t,k ) such that r t +2 4, k 1 and 2 t t 1. As above we only need Lemmas 5 and 6 (case b r,t,k b r,t 1,k and Lemma 7 (case b r,t,k <b r,t 1,k ). (d) Here we assume k = 2 and show the modifications needed to handle this case. If b r,t,1 b r,t,2, then we only need to avoid the use of the inequality a r,t,k 1 2b r,t,k 1 b r,t,2 b r,t,1, i.e. the inequality (7) a r,t,1 b r,t,2 + b r,t,1 in all cases in which it is not true. Since b r,t,1 1 and b r,t,2 2, (7) is satisfied if a r,t,1 3. Hence from now on we assume a r,t,1 2, i.e. r t +4 (Remark 6). If a r,t,2 a r,t,1 2(b r,t,2 b r,t,1 ), then we may add in Hb r,t,2 b r,t,1 general reducible conics and a r,t,2 a r,t,1 2(b r,t,2 b r,t,1 ) general lines (their postulation inside H gives no trouble by induction on r). Thus it is sufficient to have either a r,t,2 6orb r,t,2 b r,t,1 + 1 and a r,t,2 4or2 b r,t,2 a r,t,2. If (r, t) / {(4, 2), (5, 2), (5, 3), (6, 4)}, then a r,t,2 6. If b r,t,1 = 1, i.e. if r t

Unions of linear spaces 793 is odd (Remark 6), then b r,t,2 b r,t,1 + 1; in these cases we have a r,t,2 4 (a 5,3,2 = 4). We have a 6,4,2 = 5 and b 6,4,2 = 2; hence 2 b 6,4,2 a 6,4,2 ; hence the construction works even if (r, t, k) =(6, 4, 2). Since a 4,2,2 = 3 and b 4,2,2 =0, Assertion A 4,2,2 is true by [3], Theorem 3.2, or by [4]. Now assume b r,t,1 >b r,t,2, i.e. b r,t,1 = 1 and b r,t,2 = 0, i.e. r t odd and ( r+2 2 t+2 ) 2 0 (mod 3). To apply Remark 5 we need the inequality a r,t,1 2b r,t,1 +2b r,t,1 4 (i.e. the number of points Y E plus the length of the tangent vectors in Y H), i.e. a r,t,1 4, i.e. r t + 7. It remains only the cases r = t + 3 and r = t +5 with (in the latter case) t 2 (mod 3). Let F P r be the general union of A, a r,t,1 lines and one point, P. Fix a general 3-dimensional linear space H P P r containing P. In H we add the union Ea r,t,2 a r,t,1 2 general lines and a reducible conic U contained in the plane H P H and with P as its singolar point. Set X := F E χ HP (P ). The scheme X is a flat limit of a flat family whose general element is the disjoint union of A and a r,t,2 lines (Remark 2). Since Res H (X) =F, while by induction we have h 1 (H, I X H (2)) = 0, we are done. Alternatively, we could use the last part of Remark 5. Lemma 2. Fix integers r, t, k such that r t +2 4 and k 1. Assume A r,t,k. Let X P r be a general union of a t-dimensional linear subspace and a r,t,k lines. Then h 0 (I X (k)) = 0. Proof. If b t,r,k = 0, then the statement of the lemma is just the definition of A r,t,k. Assume b r,t,x > 0. Take a solution A B D of A r,t,k. Thus h 0 (I A B D (k)) = 0. For each P Sing(D) fix a 3-dimensional linear subspace H P of P r containing the connected component of D with P as its singular point. Set Y := A B D ( χ P Sing(D) H P (P )). Since A B D Y, h 0 (I Y (k)) = 0. Since Y is a flat degeneration of a flat family of general unions of a t-dimensional linear space and a r,t,k disjoint lines (Remark 2), the lemma follows from semicontinuity. Proof of Proposition 1. Since the case k = 1 is obvious, we may assume k 2. The triple (n, m, y) has critical value k. The definition of critical value and (4) gives y>a n,m,k 1. We first check that h 0 (I X (k 1)) = 0. Let Y P n be a general union of A and a n,m,k 1 lines. Since A n,m,k 1 is true (Lemma 1), h 0 (I Y (k 1)) = 0 (Lemma 2). Take as X the union of Y and y a n,m,k general lines. To prove h 0 (I X (k 1)) = 0 we just modify the proof of Lemma 1, case (a2). Let A B D be a solution of A n,m,k 1. For each P Sing(D) let H P be a general 3-dimensional linear subspace of P n containing the connected component of D with P as its singular point. Set χ := P Sing(D) χ H P (P ). Let E H be a general union of y a n,m,k 1 lines. Take as X a deformation of X := A B D χ E. To get h 1 (H, I X H(k)) = 0 we need to apply b n,m,k 1 times Lemma 5, i.e. we need the inequality a n,m,k 1 2b n,m,k 1 k 1. This inequality is true (Lemma 7). Now we fix a line L H such that A D. Since A is not contained in H, H D is a point.

794 E. Ballico For all integers r, t, k such that r t +2 4 and k 1 we define the following assertion A r,t,k r t +2. Assertion B r,t,k, r t +2 4, k 2, (r, t, k) (4, 2, 2): There is a disjoint union Y = A D P r of the t-dimensional linear subspace A, a r,t,k 2b r,t,k disjoint lines and b r,t,k reducible conics with their singular points contained in H such that L Y, L intersectsa and k lines of B and h i (I Y (k))=0,i =0, 1. Of course, B r,t,k may be true only if a r,t,k 2b r,t,k k. See Lemma 7 for this inequality. Take Y as in B r,t,k. Since h 1 (I Y (k)) = 0, Castelnuovo-Mumford s lemma implies that the homogeneous ideal of Y is generated by forms of degree at most k + 1. Since L Y and (Y L) k + 1 (remember that L A ), we get (Y L) =k +1. Lemma 3. Fix integers a b +2 3, k 2 and y 0 such that k + b k + a (8) + k +(k +1)y b a Let A P a be a b-dimensional linear subspace and L P a be a line such that L A and L A. Let E P a be a general union of y lines. Then h 1 (I A L E(k)) = 0. Proof. The lemma may be proved as Proposition 1 defining modifications of the Assertions A a,b,k in which we take A L instead of A. We prefer to give the following proof (if a b + 2. First assume a b + 3. Take a hyperplane H of P a containing A, but not containing L. For the case k 2 we start in critical value k 1 with Y = A L Y, with Y disjoint union of lines and reducible conics with their singular points contained in H. Then we add in H as in the two cases of the proof of Lemma 1 or we may just quote Lemma 1inH. Quoting Lemma 1 we do not need to use induction on b or on a. This approach does not work if a = b + 2, because in this case every line of H intersects A. For arbitrary (a, b) we may forget L and use Lemma 1 up to the critical value k 1 and in the last step add the line L; here the hyperplane H does not contain A. At the last step we add L in H passing through one of the point of A H. Here we need induction on b. The starting case of the induction, b = 1, is obvious, because in this case A L is a reducuble conic (see H k,a of [5] if a 4 ([5], p. 188). The case a = 3 may be proved as in [5], case of P 3, proof of Proposition 2.4, creating a reducible conic at the last step, one component added in the step from critical value k 2 to critical value k and meeting a degree 1 connected component of the curve with critical value k 2. Lemma 4. Fix integers r, t, k such that r t +2 4, k 2 and (r, t, k) (4, 2, 2). Then B r,t,k is true. Proof. For the case k = 2, see step (d) of the proof of Lemma 2. Now assume k 3. We copy the proof of Lemma 2, except that at each step exactly

Unions of linear spaces 795 one line of E not containing a point of B H intersects L. To satisfies this additional condition we need the inequality a 4,2,k a 4,2,k 1 b 4,2,k b 4,2,k 1 +1 and a 4,2,k 1 2b 4,2,k 1 b 4,2,k b 4,2,k 1 +1ifb 4,2,k 1 b 4,2,k (i.e. inequality stronger by +1) of the ones used in the proof of Lemma 2. These inequalities are true by Lemmas 5 and 6. Now assume b 4,2,k 1 >b 4,2,k. In this case we only need a r,t,k 1 2b r,t,k 1. This inequality is true by Lemma 7. Thus in both cases it is sufficient to prove h i (H, I X H (k)) = 0. Here the additional problem is that k + 1 points of X H are contained in L. Hence L is contained in the base locus of the linear system I X H (k) on H. We could adapt Assertions H n,n and H n,n of [5], but we prefer to use the following trick. In H we add D, E and (if b 4,2,k 1 >b 4,2,k ) χ. The fact that k lines of B intersects B H gives no trouble to the Hilbert function of (X H) D, because the points of B H not on D and the unreduced zero-dimensional components of X H are general. First assume b 4,2,k 1 b 4,2,k. In this case it is sufficient to prove h 1 (H, I (A H) D E (k)) = 0. (A H) D is a general union of a (t 1)- dimensional linear space, one line, D, intersecting it but not contained in it and a r,t,k a r,t 1,k general lines. Since (5) gives ( t 1) +k +(k +1)(ar,t,k a r,t 1,k ) ( k+r 1 r 1 ), i.e. h 0 ((A H) D E,O (A H) D E (k)) h 0 (H, O H (k)), it is sufficient to apply Lemma 3. Now assume b 4,2,k 1 >b 4,2,k. We may again apply Lemma 3. To apply Remark 5 we use Lemma 7. Proof of Theorem 1. Let Y = A B D P n be a solution of A n,k,s+1. Take any W B such that ((A W ) L) =s + 1. This is possible, because y s. Set X := A W. X is the disjoint union of A and y lines. Since X Y, h 1 (I X (t)) = 0 for all t s + 1. Since L X and (X L) =s +2, h 1 (I X (t)) s +1 t for all t>0 (Remark 1). We need to find W such that the opposite inequality holds for all t such that k t s. The curve Y is obtained in s steps from A adding disjoint lines or line linked to a previous line, adding some scheme χ HP (P ) and then making a flat deformation. It is sufficient to take as W lines introduced in the first k 1 steps, among them all the lines intersecting L, except exactly s k and in each if the last s k steps only the line intersecting L. In all deformations we may keep fixed both A and each point of intersection with L. Proof of Theorem 2. Let Y = A B D P n be a solution of A n,k,k 1. For each P Sing(D) we fix a 3-dimensional linear space containing the connected component of D containing P. Set χ := P Sing(D) χ H P (P ). Set A := A H. (a) Here we prove the statement concerning h 1. Fix an integer z such that (9) m + k 1 k + n 1 +(k +1)z m 1 n 1

796 E. Ballico m + k 2 k + n 2 (10) + kz + k(3e +2k) m 1 n 1 By the inductive assumption applied to the pair (n,m )=(n 1,m 1) there are Y 1 = A Y 2 H (with Y 2 ia disjoint union of z lines) and e disjoint lines L i H, 1 i e, such that (Y 1 L i )=k + 2 for all i, h 1 (I Y1 (k)) = e, h 1 (I Y1 (t)) = 0 for all t>kand h 1 (I Y1 (L 1 L e)(k)) = e. Let W be any union A and of some of the components of Y 1. We may assume Y 2 Y =. Obviously h 1 (H, I W (t)) h 1 (H, I Y1 (t)) for all t Z. Hence h 1 (H, I W (t)) = 0 for all t>k and h 1 (H, I W (k) e. Assume Y 1 (L 1 L e )=W (L 1 L e ). Since h 1 (I Y1 (L 1 L e)(k)) = e, we have h 1 (H, I W (k)) = e. Let E Y 2 be any union y a n,t,k 1 lines containing all lines in Y 2 intersecting one of the lines L 1,...,L e. We may find E, because y a n,t,k 1 ke by (2). Set X := Y χ E. The scheme X is a flat limit of a flat family of disjoint unions of A and y lines (Remark 2). Since Res H (X )=Y, we have h 0 (I X (k 1)) = 0. From Remark 3 and a property of Y 2 we have h 1 (I X (t)) = 0 for all t>0. By semicontinuity this property is inherited by any nearby scheme. The last part of Remark 3 and a property of Y 2 give h 1 (I X (k)) = e. This condition is preserved if we only consider smoothings of X which keep fixed E, because for any such smoothing X we have h 1 (I X (k)) e due to the presence of the set E (L 1, L e ). To start the induction we need the omitted case (n,m )=(3, 1). See [1], Proposition 1.1, for a non asymptotic statement. At each inductive step we need to check some numerical inequalities. This are always true by the case β arbitrary of Lemmas 6 and 8. (b) Here we prove the statement concerning h 0. Look at part (a). Since X contains Y and h 0 (I Y (k 1)) = 0, we have h 0 (I X (k 1)) = 0. 4. Numerical lemmas Remark 6. Fix integers r t+2 4. We have a r,t,1 = (r t)/2, b r,t,1 =0if r t is even and b r,t,1 =1ifr tis odd. We have b r,t,2 ( r+2 2 t+2 ) 2 (mod 3). We have (a 4,2,2,b 4,2,2 )=(3, 0), (a 6,4,2,b 6,4,2 )=(5, 2), (a 5,3,2,b 5,3,2 )=(4, 1), (a 5,2,2,b 5,2,2 )=(5, 0), (a 4,2,4,b 4,2,4 ) = (11, 0). Lemma 5. Fix an integer β 0. There is an integer k β 3 with the following property. Fix integers r, t, k such that r t +2 4 and k k β. Assume b r,t,k 1 b r,t,k. Then a r,t,k 1 2b r,t,k 1 b r,t,k b r,t,k 1 + β. We may take k 0 =3and k 1 =4.If(r, t) (4, 2), then we may take k 1 =3. Proof. Since 0 b r,t,k 1 k 1 and b r,t,k k and ka r,t,k 1 ( r+k 1 r t+k 1 ) k 1 the lemma is true if r + k 1 t + k 1 (11) k(2k 1+β)+1 k r t The inequality (6) is true if either β =0,k = 3 and (r, t) (2, 2) or β {0, 1} and k 4. Since r t +2 4, for arbitrary β (11) is true if k 0 and if k 2 and the case (r, t) =(4, 2) is true for that integer k, then the same

Unions of linear spaces 797 integer k works for arbitrary (r, t) such that r t +2 4. Now assume (r, t, k) =(4, 2, 3). Since a 4,2,2 =3,b 4,2,2 = 0 and b 4,2,3 = 3, the inequality in the statement of the lemma is true if and only if β =0. Lemma 6. Fix an integer β 0. There is an integer k β 2 with the following property. Fix integers r, t, k such that r t +2 4 and k k β. Assume b r,t,k 1 b r,t,k. Then a r,t,k a r,t,k 1 b r,t,k b r,t,k 1 + β +1. We may take k 0 =2. Proof. Assume a r,t,k a r,t,k 1 b r,t,k b r,t,k 1 + β. From (5) we get t + k 1 r + k 1 (12) + k(b r,t,k b r,t,k 1 )+(k +1)β + a r,t,k 1 m 1 r 1 We have b r,t,k b r,t,k 1 k with strict inequality if b r,t,k 1 > 0. Hence from (4) for k := k 1 and (12) we get (13) t + k 1 r + k 1 t + k 1 r + k 1 k + + k 3 + k(k +1)β k t 1 r 1 m r 1 If β = 0, then (13) fails, unless (r, t, k) =(4, 2, 2). Since a 4,2,1 =1,b 4,2,1 =0, a 4,2,2 = 3 and b 4,2,2 = 0, Lemma 6 is true if β = 0. Now assume β>0. Since r t +2 4, (13) fails if k 0 and this failure arises for all k 2 for which (13) fails for the pair (r, t) :=(4, 2). We may take as k β this large integer. Lemma 7. For all integers r, t, k such that r t +2 4 and k 2 we have a r,t,k 2b r,t,k k, unless (r, t, k) =(6, 4, 2). We have (a 6,4,2,b 6,4,2 )=(5, 2). Proof. Assume a r,t,k 2b r,t,k k 1. Hence a r,t,k 3k 1. From (4) we get t + k r + k (14) +(k + 1)(3k 1) t r If (r, t) =(4, 2), then (14) fails if and only if k 5. In the case (r, t, k) = (4, 2, 4) the left hand side and the right hand side of (14) are equal; in this case we use b 4,2,4 = 0 and a 4,2,4 = 11. If r = t + 2 and t 3, then (14) fails for all k 3. If r t + 3, then (14) fails for all k 3; if k = 2 it fails if and only if (r, t) (5, 2). Now assume k =2. If(r, t) / {(4, 2), (5, 2), (5, 3), (6, 4)}, then a r,t,2 6. Hence the lemma is true for k =2if (r, t) / {(4, 2), (5, 2), (5, 3), (6, 4)}. Since (a 4,2,2,b 4,2,2 )=(3, 0), (a 6,4,2,b 6,4,2 )= (5, 2), (a 5,3,2,b 5,3,2 )=(4, 1), (a 5,2,2,b 5,2,2 )=(5, 0), we are done. As in lemmas 6 and 7 we get the following lemma. Lemma 8. Fix an integer β 0. There is an integer k β 2 with the following property. Fix integers r, t, k such that r t +2 4 and k k β. Then a r,t,k a r,t,k 1 k + β.

798 E. Ballico References [1] E. Ballico, On smooth curves with multisecant lines. Rev. Roumaine Math. Pures Appl. 33 (1988), no. 4, 267 274. [2] E. Ballico, Postulation of disjoint unions of lines and a few planes, preprint. [3] E. Carlini, M. V. Catalisano and A. V. Geramita, Subspace arrangements, configurations of linear spaces and quadrics containing them, arxiv:0909.3821[math.ag] 21 Sep 2009. [4] E. Carlini, M. V. Catalisano and A. V. Geramita, Bipolynomial Hilbert functions, arxiv:0910.3569[math.ag]. [5] R. Hartshorne and A. Hirschowitz, Droites en position générale dans P n, Algebraic Geometry, Proceedings, La Rábida 1981, 169 188, Lect. Notes in Math. 961, Springer, Berlin, 1982. Received: October, 2009